# Jacobian matrix with 2 variables !

1. Feb 8, 2007

### feelau

jacobian matrix with 2 variables plz help!

1. The problem statement, all variables and given/known data
so we have z=x^2+x^3 and z=y+sin(x). Find the jacobian matrix of this system. Find the determinant of this jacobian.

3. The attempt at a solution
The determinant part is easy, the only problem is trying to set this up. I'm not sure where to start. I think I need to take partial derivatives....do we do like dz/dx and dz/dy and just put them into matrix? plz help

2. Feb 8, 2007

### HallsofIvy

Staff Emeritus
Yes, exactly. The partial derivatives of the first function, with respect to x and y, are the first row of the matrix, the first derivatives of the second function, with respect to x and y, are the second row of the matrix.

Just out of curiosity, how is the determinant part "easy" if you don't know what you are to take the determinant of?

3. Feb 8, 2007

### feelau

what do you mean? the determinant doesn't necessarily have to be a certain numerical value, just show what the answer is symbolically(that's what I believe atleast heh). so if I did this right the determinant should be just 2x+3x^2

4. Feb 9, 2007

### joob

Are you sure you wrote the problem out right? both equations are for z?

5. Feb 9, 2007

### HallsofIvy

Staff Emeritus
Yes, that is correct. I said nothing about "a certain numerical value". I'm still wondering how you got 2x+ 3x2 for the determinant of the Jacobean matrix if you did not know what that matrix was! Oh, and I assumed that the two "z"s actually referred to different variables. That was what joob was complaining about. Normally, a Jacobean matrix reflects a change in variables. changing x,y to u, v or z,w would make sense. Changing x,y to z, z would not!

6. May 21, 2008

### Ultimâ

$$z_1=x^2+x^3 \medskip z_2=y+sin(x)$$

Jacobian is:
$$J=\left[ \begin{array}{cc} \frac{dz_1}{dx} & \frac{dz_1}{dy} \\ \frac{dz_2}{dx} & \frac{dz_2}{dy} \end{array} \right]$$

...which is
$$J=\left[ \begin{array}{cc} 2x & 3x^2 \\ cos(x) & 1 \end{array} \right]$$

and determinant:

$$2x -3x^2cos(x)$$